Réitigh f(x)=x^9/x^2-6x+5

$$f ( x ) = \frac { x ^ { 9 } } { x ^ { 2 } - 6 x + 5 }$$

Difreálaigh w.r.t. x

$$\frac{\left(7x^{2}-48x+45\right)x^{8}}{x^{4}-12x^{3}+46x^{2}-60x+25}$$

Céimeanna ag Baint Úsáid as Riail na nDíorthach don Líon

Do dhá fheidhm indifreáilte ar bith, is ionann díorthach líon an dá fheidhme agus an t-ainmneoir méadaithe faoi dhíorthach an uimhreora lúide an t-uimhreoir méadaithe faoi dhíorthach an ainmneora, agus iad ar fad roinnte faoin ainmneoir cearnaithe.

$$\frac{\left(x^{2}-6x^{1}+5\right)\frac{\mathrm{d}}{\mathrm{d}x}(x^{9})-x^{9}\frac{\mathrm{d}}{\mathrm{d}x}(x^{2}-6x^{1}+5)}{\left(x^{2}-6x^{1}+5\right)^{2}}$$

Is ionann díorthach iltéarmaigh agus suim dhíorthaigh a théarmaí. Is ionann díorthach téarma thairisigh agus 0. Is ionann díorthach ax^{n} agus nax^{n-1}.

$$\frac{\left(x^{2}-6x^{1}+5\right)\times 9x^{9-1}-x^{9}\left(2x^{2-1}-6x^{1-1}\right)}{\left(x^{2}-6x^{1}+5\right)^{2}}$$

Simpligh.

$$\frac{\left(x^{2}-6x^{1}+5\right)\times 9x^{8}-x^{9}\left(2x^{1}-6x^{0}\right)}{\left(x^{2}-6x^{1}+5\right)^{2}}$$

Méadaigh x^{2}-6x^{1}+5 faoi 9x^{8}.

$$\frac{x^{2}\times 9x^{8}-6x^{1}\times 9x^{8}+5\times 9x^{8}-x^{9}\left(2x^{1}-6x^{0}\right)}{\left(x^{2}-6x^{1}+5\right)^{2}}$$

Méadaigh x^{9} faoi 2x^{1}-6x^{0}.

$$\frac{x^{2}\times 9x^{8}-6x^{1}\times 9x^{8}+5\times 9x^{8}-\left(x^{9}\times 2x^{1}+x^{9}\left(-6\right)x^{0}\right)}{\left(x^{2}-6x^{1}+5\right)^{2}}$$

Chun cumhachtaí an bhoinn chéanna a mhéadú, suimigh a n-easpónaint.

$$\frac{9x^{2+8}-6\times 9x^{1+8}+5\times 9x^{8}-\left(2x^{9+1}-6x^{9}\right)}{\left(x^{2}-6x^{1}+5\right)^{2}}$$

Simpligh.

$$\frac{9x^{10}-54x^{9}+45x^{8}-\left(2x^{10}-6x^{9}\right)}{\left(x^{2}-6x^{1}+5\right)^{2}}$$

Cuir téarmaí cosúla le chéile.

$$\frac{7x^{10}-48x^{9}+45x^{8}}{\left(x^{2}-6x^{1}+5\right)^{2}}$$

Do théarma ar bith t, t^{1}=t.

$$\frac{7x^{10}-48x^{9}+45x^{8}}{\left(x^{2}-6x+5\right)^{2}}$$

Luacháil

$$\frac{x^{9}}{\left(x-5\right)\left(x-1\right)}$$